Find the value of alpha and theta?

Z_1=2(cos(18^@)+isin(18^@))
Z_2=alpha(cos(40^@)+isin(40^@))
Z_3=32(cos(theta)+isin(theta))
Find value of alpha and theta if
(Z_1^10)/(Z_2-Z_3)=4i

1 Answer
Jan 19, 2018

Given:

Z_1=2(cos(18^@)+isin(18^@))" [1]"

Z_2=alpha(cos(40^@)+isin(40^@))" [2]"

Z_3=32(cos(theta)+isin(theta))" [3]"

(Z_1^10)/(Z_2-Z_3)=4i" [4]"

Substitute equation [1] into equation [4]:

((2(cos(18^@)+isin(18^@)))^10)/(Z_2-Z_3)=4i

Use De Moivre's formula:

(2^10(cos(18^@xx10)+isin(18^@xx10)))/(Z_2-Z_3)=4i

(1024(cos(180^@)+isin(180^@)))/(Z_2-Z_3)=4i

(1024(-1+i0))/(Z_2-Z_3)=4i

(-1024+i0)/(Z_2-Z_3)=4i" [4.1]"

Multiply both sides of equation [4.1] by Z_3-Z_2:

1024+i0=4i(Z_3-Z_2)" [4.2]"

Substitute equations [2] and [3] into equation [4.2]:

1024+i0=4i(32(cos(theta)+isin(theta))-alpha(cos(40^@)+isin(40^@)))

Perform the multiplication:

1024+i0=4i(32cos(theta)+i32sin(theta))-alphacos(40^@)-ialphasin(40^@))

1024+i0=i128cos(theta)-128sin(theta)-i4alphacos(40^@)+4alphasin(40^@))

Separating the real and imaginary parts into two equations:

1024 = 4alphasin(40^@)-128sin(theta)" [5]"

i0 = i128cos(theta)-i4alphacos(40^@)" [6]"

Solve equation [6] for alpha in terms of theta:

0 = 32cos(theta)-alphacos(40^@)

alpha = 32cos(theta)/cos(40^@)" [6.1]"

Substitute equation [6.1] into equation [5]:

1024 = 4(32cos(theta)/cos(40^@))sin(40^@)-128sin(theta)

Multiply both sides by cos(40^@)/128:

8cos(40^@) = cos(theta)sin(40^@)-sin(theta)cos(40^@)

Multiply both sides by -1:

-8cos(40^@) = sin(theta)cos(40^@)-cos(theta)sin(40^@)

The right side is the identity for sin(A-B) where A=theta and B = 40^@:

sin(theta-40^@) = -8cos(40^@)

The above equation does not have a solution, because the right side is outside of the the range of the sine function, -1 <= sin(x) <= 1. This means that there are not real values for theta and alpha